355 – Mechatronic System Part II - AC Drives 2
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Transcript of 355 – Mechatronic System Part II - AC Drives 2
EEM EEM 355 –Mechatronic SystemMechatronic System
AC DRIVES 2BY:
DR. ROSMIWATI MOHD MOKHTAR
Example 1
A 15 hp, 460 V, 60 Hz, 3-phase squirrel cage5 p, 4 , , 3 p q ginduction motor is having a torque speed curve givenas below.
By neglecting windage and friction losses, calculate the y g g g ,power supplied to the rotor when the machine runs
a) As a motor at 1650 rpmb) As a brake at 750 rpmc) As a generator at 2550 rpm
Motor: nm = +1650 rpm; m = +100 Nm
55.9
m
mm
nn
P
083.01800
16501800
s
ms
nnns
91001650
55.9
mmm
nP
kW851828.17
1
PP
PsPm
rm
kW 28.1755.9
kW85.18
083.011 sPr
Brake: nm = -750 rpm; m = +60 Nm
55.9
m
mm
nn
P
417.11800
7501800
s
ms
nnns
960750
55.9
mmm
nP
kW31171.4
1
PP
PsPm
rm
kW 71.455.9
kW3.11
417.111
sPr
Generator: nm = +2550 rpm; m = ??
Example 2
A 3-phase, 4-pole, 1800 rpm squirrel cage induction motori d 8 V 6 H W his rated at 208 V, 60 Hz. We want the motor to turn at a no-load speed of about 225 rpm while maintaining the sameflux in the air gap. Calculate the required voltage andf b li d hfrequency to be applied to the stator.
To run at 225 rpm, the voltage and frequency must beTo run at 225 rpm, the voltage and frequency must bereduced in the ratio of 225/1800 = 0.125E
V l i d 8 6 V Voltage required = 0.125 x 208 = 26 V
and frequency = 0.125 x 60 = 7.5 Hzand frequency 0.125 x 60 7.5 Hz
Kinetic Energy
Recall that the dynamic braking will stop motor bydissipating its stored kinetic energy into resistive load.
The kinetic energy is a form of mechanical energy is givenby the equationby the equation
2322 1048.521
21 JnmvJKE
22J = moment of inertia [kg.m2]= rotational speed [rad/s]m mass of the body [kg]m = mass of the body [kg]v = translational speed [m/s]n = rotational speed [rpm]5.48 x 10-3 = constant to take care of units [exact value = 2/1800]
Torque, Inertia and Change in Speed
The rate of change of speed depends upon theg p p pinertia, as well as on the torque. The equation thatrelates these factors is
Jtn
55.9
Jn = change in speed [rpm]= Torque [Nm]t = Interval of time during which the
torque is applied [s]torque is applied [s]J = moment of inertia [kg.m2]9.55 = constant to take care of units
[exact value = 30/]
Example 3
The flywheel having total moment of inertia 10.6y gkg.m2 turns at 60 rpm. We wish to increase its speedto 600 rpm by applying a torque of 20 Nm. For howl t th t b li d?long must the torque be applied?
55.9
tn
559
nJtJ
n
302055.9
6.106060055.9
s30
Electronic AC Drives
Static frequency changersConvert the incoming line frequency directly into the desired loado Convert the incoming line frequency directly into the desired loadfrequency.
o Cycloconverters fall into this category.o Used to drive synchronous and squirrel-cage induction motors.o Used to drive synchronous and squirrel cage induction motors.o Motor can be started, stopped, reversed ad decelerated by
regenerative braking.o The stator voltage is automatically adjusted in relation to the
frequency to maintain a constant flux in the machine.o To obtain regenerative braking,
frequency produced bycycloconverter must be slightlyless than the frequencycorresponding to the speed of themotor.
Variable speed drive using cycloconverter
Squirrel-cage induction motor fed from a 3-phase cycloconverter
Static voltage controllersgo Enable speed and torque control by varying the ac voltage.o Used with squirrel cage induction motors.
S i l ll l d f i d io Static voltage controllers are also used to soft-start inductionmotors.
o Particularly useful for a motor driving a blower or centrifugaly g gpump.
Variable speed drive using static switch
Variable-speed blower motor
Variable-voltage speed control of a squirrel-cage induction motor using back-to-back thyristors
Rectifier-inverter systems with line commutation- Rectify the incoming line frequency to dc, and the dc is reconverted to
ac by an inverter.- The inverter is line commutated by the motor it drives.y- Used to control speed of wound-rotor induction motors.- Adding a chopper to the rectifier circuit also enable for control of speed
of wound-rotor induction motorof wound rotor induction motor.
Rectifier-inverter systems with self commutation- Rectify the incoming line frequency to dc, and the dc is reconverted to
ac by an inverter.- The inverter is self commutated, generating its own frequency., g g q y- Used to control speed of squirrel cage induction motors.
Current-fed frequency converter
Voltage-fed frequency converter
Pulse-width modulation systemsy- Relatively new in industry.- Enable variable speed induction drives ranging from p g g
zero speed and up.
Example 4
A 3-phase, squirrel-cage induction motor has a fullload rating of 25 hp, 480 V, 1760 rpm, 60 Hz. The 3independent windings each carry a rated current of20 A. The cycloconverter is connected to a 3-phase,20 A. The cycloconverter is connected to a 3 phase,60 Hz line and generates a frequency of 8 Hz.Calculate the approximate value of the following:
) h ff i l h i dia) The effective voltage across each windingb) The no-load speed) Th d t t d tc) The speed at rated torque
d) The effective current in the windings at ratedtorquetorque
Answer 4
a) The flux in the motor should remain the same at all frequencies.C l f f f 8 H h l h i diConsequently, for a frequency of 8 Hz, the voltage across the windingsmust be reduced in proportion. Thus, the effective voltage across eachwinding is
V 64480608
E
b) The full load speed at 60 Hz = 1760 rpm. Consequently, this is a 4-pole motor. Thus, its synchronous speed is
60120120 f
The no load speed at 8 Hz is
rpm 18004
60120120
pfns
The no-load speed at 8 Hz isrpm 2401800
608
n
c) When the motor is operating at 60 Hz, the slip speed at rated torque iis
4017601800
msslip nnn
Consequently, the slip is again 40 rpm when the motor developsrated torque at 8 Hz. Therefore, the speed at rated torque
rpm40
q , p q
d) Because the flux in the motor is the same at 8 Hz and 60 Hz it
rpm 20040240 n
d) Because the flux in the motor is the same at 8 Hz and 60 Hz, itfollows that rated torque will be developed when the current in thestator windings reaches its rated value, namely 20 A. Effective current in the windings at rated torque = 20 A Effective current in the windings at rated torque = 20 A.
Example 5
A Y-connected squirrel-cage induction motor has thefollowing ratings and parameters:
400 V, 50 Hz, 4-pole, 1370 rpm, Rs = 2 , Rr’=3 , Xs = Xr’ = 3.5
Motor is controlled by a voltage source inverter at constantV/f ratio. Inverter allows frequency variation from 10 to 50/ q y 5Hz. Determine,a) Speed for the frequency of 30 Hz and 80% of full load
ttorqueb) Frequency for a speed of 1000 rpm and full load torque
a) The speed at 50 Hz = 1370 rpm. Consequently, this is a 4-pole motor.Thus its synchronous speed at 50 Hz isThus, its synchronous speed at 50 Hz is
rpm 15004
50120120
pfns
At 50 Hz, drop in speed from no-load to full load torque,= 1500 – 1370 = 130 rpm.
p
5 37 3 p
Drop in speed from no-load to 80% of full load torque,= 0.8 x 130 = 104 rpm.
Synchronous speed at 30 Hz = 30/50 x 1500 = 900 rpm.
Motor speed for frequency 30 Hz and having 80% of full load torque =900-104 = 796 rpm
b) Frequency for a speed of 1000 rpm and full load torqueb) Frequency for a speed of 1000 rpm and full load torque
Drop in speed from no load to full load torque = 130 rpm.
Synchronous speed = 1000 + 130 = 1130 rpm.
Frequency,
120 f
Hz673711304
120
s
s
pnf
pfn
Hz 67.37120120
f
Let say the regenerative braking operation isy g g pemployed, determine;
a) Speed for the frequency of 30 Hz and 80% of fullload torque
b) Frequency for a speed of 1000 rpm and full loadtorque
c) What will be the answer to (a) to (b), if the drivek d d b kworks under dynamic braking?
Answer 5
a) The speed at 50 Hz = 1370 rpm. Consequently, this is a 4-pole motor.Thus, its synchronous speed at 50 Hz isThus, its synchronous speed at 50 Hz is
rpm 15004
50120120
pfns
Increase in speed from no-load to full load torque,= 1500 – 1370 = 130 rpm.
Increase in speed from no-load to 0.8 of full load torque,= 0.8 x 130 = 104 rpm.
Synchronous speed at 30 Hz = 30/50 x 1500 = 900 rpm.
M t d f f q 30 H d h i 80% f f ll l d t q Motor speed for frequency 30 Hz and having 80% of full load torque =900+104 = 1004 rpm
b) Frequency for a speed of 1000 rpm and full load torque
Synchronous speed = 1000 – 130 = 870 rpm.
Frequency,120
s pfn
Hz 29120
8704120
spnf
c) In both regenerative and dynamic braking, motor works as agenerator. The two braking methods only differ in the way brakingenergy is disposed off (in former it is transferred to the source and inlatter it is dissipated in a resistor) Hence answer will be the samelatter it is dissipated in a resistor). Hence, answer will be the same.